Here is an odd application of multiple integration. I know that \(e^{-x^2}\) has no elementary antiderivative. Therefore, the intergal
\begin{equation*}
A = \int_{-\infty}^\infty e^{-x^2} dx
\end{equation*}
cannot be evaluated directly. However, this is a very important integral: \(e^{-x^2}\) is the normal distribution and, in Statistics, I need to integrate it frequently. I’ll use integrals over \(\RR^2\text{,}\) strangely enough, to calculate this integral by squaring the single variable integral.
\begin{align*}
A \amp = 2 \int_{0}^\infty e^{-x^2} dx\\
A^2 \amp = \left( 2 \int_{0}^\infty e^{-x^2} dx \right)
\left(2 \int_{0}^\infty e^{-y^2} dy \right)
\end{align*}
The second integral uses a new variables since variables of integration only matter inside the integral. Then I can combine the two integrals.
\begin{align*}
A^2 \amp = 4 \int_0^\infty \int_0^\infty e^{-x^2} e^{-y^2}
dx dy\\
\amp = 4 \int_0^\infty \int_0^\infty e^{-(x^2 + y^2)} dx
dy
\end{align*}
Now I am going to do a substitution in the \(y\) variable. Treating the \(x\) variable as a constant, I replace \(y\) with \(y = xs\) so that \(dy = x ds\text{.}\) If \(y=0\) then \(s=0\) and as \(y \rightarrow
\infty\text{,}\) \(s \rightarrow \infty\text{,}\) so the bounds for \(s\) remain the same as the bounds for \(y\text{.}\) Remember, \(x\) is a constant through this whole substitution. (Using Fubini’s Theorem, I could think of \(y\) as the inside integral for this substitution, and then switch back to thinking of \(x\) as the inside integral after the substitution.)
\begin{align*}
A^2 \amp = 4 \int_0^\infty \int_0^\infty e^{-x^2 (1+s^2)}
x dx ds\\
\amp = 4 \int_0^\infty \left. \frac{-1}{2(1+s^2)}
e^{-x^2(1+s^2)} \right|_0^\infty ds\\
\amp = 2 \int_0^\infty \frac{1}{1+s^2} ds\\
\amp = \lim_{a \rightarrow \infty} 2 \arctan a - \arctan 0
= \frac{2\pi}{2} = \pi\\
A \amp = \int_{-\infty}^\infty e^{-x^2} dx = \sqrt{\pi}
\end{align*}
I recover the area under the bell curve: \(\sqrt{\pi}\text{.}\) It’s a very strange result. However, if you taken any statistics and worked on normal distributions, likely you will recall the presence of these strange \(\sqrt{\pi}\) terms. Now I know they are present to normalize the area (since we want a probability function to have area one under its graph).
